Help deriving elastic collisions equations?

Question

How do I turn the equation, m1v0+m2v0=m1v1f+m2v2f into the equation v1f= [(m1-m2)/(m1+m2)]v0.

My Physics teacher said it's possible and that it involves "messy" algebra. This is a problem my teacher wouldn't help me with and I've been trying for a couple hours and am stuck. Can you please show me step by step.

Answer 1

You can't do this. (Or, at least, you can but it's only correct if m2 = 0.) Note that in an elastic collision if both objects have the same initial velocity they will both keep that velocity.

I think the original equation is supposed to be
m1 v0 = m1 v1f + m2 v2f
In other words, the second object is stationary before the collision.

In this case, given that it is an elastic collision we also know that the kinetic energy has not changed:
1/2 m1 v0^2 = 1/2 m1 v1f^2 + 1/2 m2 v2f^2
=> m1 v0^2 = m1 v1f^2 + m2 v2f^2
=> m1 (v1f^2 - v0^2) + m2 v2f^2 = 0
=> m1 m2 (v1f^2 - v0^2) + m2^2 v2f^2 = 0
From the original equation we can write m2 v2f = m1 (v1f - v0), so we get
m1 m2 (v1f^2 - v0^2) + m1^2 (v1f - v0)^2 = 0
=> m1 m2 (v1f - v0) (v1f + v0) + m1^2 (v1f - v0)^2 = 0
Dividing through by m1 (v1f - v0):
=> m2 (v1f + v0) + m1 (v1f - v0) = 0
=> (m2 + m1) v1f + m2 v0 - m1 v0 = 0
=> v1f (m1 + m2) = v0 (m1 - m2)
=> v1f = v0 (m1 - m2) / (m1 + m2) as you had.

Noet that for the case where the objects have equal but opposite velocities before the collision, the momentum equation is
m1 v0 - m2 v0 = m1 v1f + m2 v2f
and we get
v1f = v0 (m1 - 3m2) / (m1 + m2).

In general, if the first object has initial velocity v0 and the second has initial velocity kv0, you will get
v1f = v0 (m1 + (2k-1) m2) / (m1 + m2).
Note that if k = 1 (objects have same velocity and direction) you get v1f = v0 (no change) as expected. Also, note that if the masses are equal this simplifies to v1f = kv0 (and hence v2f = v0), i.e. they exchange velocities, also as expected for elastic collisions.

Answer 2

Elastic Collision Equation

Answer 3

Formula For Elastic Collision

Answer 4

You must consider two things:

a) Conservation of energy
By definition in an elastic collision no energy is lost or gained.
The total kinetic energy (1/2 m v2) before the collision equals the total kinetic enery after the collision.
Write this down in the form of an equation:



b) Conservation of Momentum
In a collision the total momentum before the collision equals the total momentum after the collision.
You have almost correctly written down the equation for this:
(m1v1)B+(m2v2)B=(m1v1)A+(m2v2)A


Now manipulate the two equations to get rid of any unknowns (letters) that you do not want in it, and re-arrange it in its final form.

Note: This is difficult to show on the WWWnet due to all the subscripts and squarefunctions getting mixed up!

Answer 5

use the chain rule to derive velocity